How do you evaluate the six trigonometric functions given t=5π/4?

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How do you evaluate the six trigonometric functions given t=5π/4?

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Answer 1 · 2018-03-16T08:43:04

As below.

Explanation:

To find 6 trigonometric functions of $\frac{5 π}{4}$ $(5pi) / 4$

$\frac{ˆ 5 π}{4}$ $hat (5pi)/4$ is $> 180^{\circ}$ $> 180^@$ & < $270^{\circ}$ $270^@$ Hence it falls in the third quadrant where only $tan θ and cot θ$ $tan theta and cot theta$ positive#

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$\frac{sin (5 π)}{4} = sin (π + \frac{π}{4}) = - sin (\frac{π}{4}) = - (\frac{1}{\sqrt{2}})$ $sin (5pi)/4 = sin (pi + pi/4) = - sin (pi/4) = color(red)(- (1/sqrt2)$

$\frac{csc (5 π)}{4} = csc (π + \frac{π}{4}) = - csc (\frac{π}{4}) = - (\sqrt{2})$ $csc (5pi)/4 = csc (pi + pi/4) = - csc (pi/4) = color(red)(- (sqrt 2)$

$\frac{cos (5 π)}{4} = cos (π + \frac{π}{4}) = - cos (\frac{π}{4}) = - (\frac{1}{\sqrt{2}})$ $cos (5pi)/4 = cos (pi + pi/4) = - cos (pi/4) =color(red)( - (1/sqrt2)$

$\frac{sec (5 π)}{4} = sec (π + \frac{π}{4}) = - sec (\frac{π}{4}) = - (\sqrt{2})$ $sec (5pi)/4 = sec (pi + pi/4) = - sec (pi/4) = color(red)(- (sqrt2)$

$\frac{tan (5 π)}{4} = tan (π + \frac{π}{4}) = tan (\frac{π}{4}) = 1$ $tan (5pi)/4 = tan (pi + pi/4) = tan (pi/4) = color(green)(1$

$\frac{cot (5 π)}{4} = cot (π + \frac{π}{4}) = cot (\frac{π}{4}) = 1$ $cot (5pi)/4 = cot (pi + pi/4) = cot (pi/4) = color(green)(1$

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How do you evaluate the six trigonometric functions given t=5π/4?

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